THE PYTHAGOREAN DISTANCE FORMULA

BASIC TO TRIGONOMETRY and calculus is the theorem that relates the squares drawn on the sides of a right-angled triangle. Credit for the proving the theorem goes to the Greek philosopher Pythagoras, who lived in the 6th century B. C.

Here is the statement of the theorem:

In a right triangle the square drawn on the side opposite the right angle is equal to the squares drawn on the sides that make the right angle.

That means that if ABC is a right triangle with the right angle at A, then the square drawn on BC opposite the right angle, is equal to the two squares together on CA, AB.

In other words, if it takes one can of paint to paint the square on BC, then it will also take exactly one can to paint the other two squares.

The side opposite the right angle is called the hypotenuse ("hy-POT'n-yoos"; which literally means stretching under).

The subscript 1 labels the coördinates of the first point; the subscript 2 labels the coördinates of the second. We write the absolute value because distance is never negative.

Here then is the Pythagorean distance formula between any two points:

It is conventional to denote the difference of
x-coördinates by the symbol Δx ("delta-x"):

Δx = x2 − x1

Similarly,

Δy = y2 − y1

Therefore,

Example 2. Calculate the distance between the points (1, 3) and (4, 8).

Solution.

Δx

=

4 − 1 = 3.

Δy

=

8 − 3 = 5.

Therefore,

Note: It does not matter which point we call the first and which the second. Alternatively,

Δx

=

1 − 4 = −3.

Δy

=

3 − 8 = −5.

But (−3)2 = 9, and (−5)2 = 25. The distance between the two points is the same.

Example 3. Calculate the distance between the points (−8, −4) and (1, 2).

Solution.

Δx

=

1 − (−8) = 1 + 8 = 9.

Δy

=

2 − (−4) = 2 + 4 = 6.

Therefore,

Problem 4. Calculate the distance between (2, 5) and (8, 1)

Problem 5. Calculate the distance between (−11, −6) and (−16, −1)

A proof of the Pythagorean theorem

Let a right triangle have sides a, b, and hypotenuse c. And let us arrange four of those triangles to form a square whose side is a + b. (Fig. 1)

Now, the area of that square is equal to the sum of the four triangles, plus the interior square whose side is c.

Two of those triangles taken together, however, are equal to a rectangle whose sides are a, b. The
area of such a rectangle is a times b: ab. Therefore the four triangles together are equal to two such
rectangles. Their area is 2ab.

As for the square whose side is c, its area is simply c2. Therefore, the area of the entire square is

c2 + 2ab. . . . . . . (1)

At the same time, an equal square with side a + b (Fig. 2) is made up of a square whose side is a, a square whose side is b, and two rectangles whose sides are a, b. Therefore the area of that square is

a2 + b2 + 2ab.

But this is equal to the square formed by the triangles, line (1):

a2 + b2 + 2ab = c2 + 2ab.

Therefore, on subtracting the two rectangles 2ab from each square, we are left with